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4.1. Introduction:
Wear is defined as the gradual loss of material from
the interacting surfaces during the relative motion under the
normal load. Wear is classified as mild wear and severe
wear. Mild wear is therefore generally associated with lowloads where metallic interactions are somewhat inhibited
and the wear debris consists of fine particles and is usually
in the form of oxides. This does not imply that metallic
contacts have never occurred at all, since the resulting
metallic debris would tend to become oxidized at the high
local temperature at the interface. Nevertheless the nature
of the surface asperity interaction is relatively gentle,
resulting in characteristically mild wear and smoothing of
the surfaces.
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Fig. 4.1 The transition phenomena in wear
At higher load a much coarser wear process occurs.
The wear debris is of a much larger particle-size, the worn
surfaces are much rougher and the increase in volume
changes the wear rate by several orders of magnitude. This
is the so-called severe wear regime. A starting fact about
these two types of wear behavior is the very rapid transition
from one mode to the other as the load is increased (fig.
4.1). In this figure it should be noted that the wear rate
suddenly changes by more than one hundred times. With
some materials at even higher loads the increasing
temperatures cause metallurgical changes in the materials
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such as to increase their inherent hardness. These effects
can then lead to a second transition from the severe wear
back to the mild wear regime.
4.2. Mechanisms of Wear:
We now consider the detailed mechanisms by which
material may be removed from the surface. The most
common mechanisms are:(a) Adhesive wear, (d) Corrosive wear.
(b) Abrasive wear, (e) Erosive wear
(c) Fatigue wear
In some situations more than one of these mechanisms
may be operative at the same time, and this is one of the
reasons for the complexity of wear studies.
4.2.1. Adhesive Wear:
When the two surfaces are having identical hardness
then the contact between surfaces occurs at the tips of the
asperities, which then deform under load. Tile nature of the
adhesion between such asperity interactions is modified by
surface films so that the metallic adhesion characterized in
the simple friction theory is somewhat modified. But as
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translation occurs these surface films are to some extent
disrupted and adhesion will occur at a certain proportion of
these contact, as can be appreciated in view of thedistribution of asperity heights discussed in chapter 2.
From these considerations and our knowledge of the
nature of the contact of rough surfaces we can now predict
a ‘wear equation’, If one assumes that the wear particles are
geometrically similar, the wear volume would be expectedto be proportional to the real areas of contact at which
adhesion occurs, and also to the distance of sliding. Since
the real area of contact for the plastic interaction of
asperities is given by
A = W / H
The wear volume V is given by
V∝ A x L ∝ W x L
H
Where L is the distance of s\liding. Thus the adhesive wear
law becomes WV = K W x L
H
Or in words the: ‘laws’ of adhesive wear are:
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(a) The volume of Wear is proportional to the distance
of sliding. This relationship has been justified by
experience for a wide range of conditions.(b) The volume of wear is proportional to the applied
load. This has also been shown to be true in many tests for
limited ranges of load, although as wear mechanisms
change with increasing load some abrupt transitions have
been observed; see fig. 4.1.(c) The volume of wear is inversely proportional to the
hardness of the softer material. This has also been shown to
be valid, particularly for pure metals.
Recalling the physical nature of the wear process
arising from adhesion we can give a physical meaning to
the constant of proportionality K. often called the adhesive
wear coefficient or the Archard constant. Surface films
together with the asperity contact forms the adhesive
contacts are only significant on adhesive wear. Since
shearing due to sliding probably formed where the
junctions are stronger than the underlying material, So that
some material from the surfaces is torn away and
eventually released by the continued sliding. So we can see
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that K may be probability factor, that is, the factor which
indicates the probability of wear particles being created by
the adhesive effect between the populations of asperities onthe two rubbing surfaces.
Fig. 4·2 Surface contact of an identical
hemispherical asperity
Theory of Adhesive Wear:
Consider two surfaces of an identical material having
the same hardness. Assume that the asperities on both the
surfaces are identical. In nature of such surfaces are in
contact then the deformed area of the asperities is a circular
area of radius 'a' under the application of load W.
Consider anyone asperity in contact when the asperity
is in relative motion the function of asperity will be shear
from the weaker bond of the asperity. Let the load carried
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by anyone asperity is W1 = W/n where n is the total
numbers of asperities in contact.
Let the particle separate out from such asperity is of ahemispherical in shape having a radius, ‘a’
The volume of material in dist. is vol. of sphere V/2 =
2/3 π r 3.
V1 = (4/3 π a3) / 2 ( = 2a)X X
V1 = 1 = 1/3 π a2 = 1/3 A1
X 3
Since πa2 is the area of contact, where is proportional to the
ratio W1/H.
Where H = hardness of material.
A1 ∝ W1
But A = A1 + A2 + …………… + An
W = W1 + W2 + …………… + Wn
& also
A1 = A2 = A1 =…………… = An for similar shape of
asperity
W1 = W2 = W1 =…………… = Wn
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A = n A1
W = n W1
A1 = A / n ∝ W1 = W / n
A ∝ W
The total volume of wear from the surface, if all the
asperities are effective in removing the material then.
V = n V1
V1 = 1 A1 = 1 W1
X 3 3 H
Therefore volume of wear for displacement is
V1 = 1 W1X3 H
V = n V1
= n . 1 W1X But nW1 = WT = W
3 H
V = W X
3 H
If wear constant K adh can be introduced for considering
the actual numbers of the asperity in contact.
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V= K adh . W / 3H.
K adh = 1, When all asperities in contact are effective in
wear = 0.1, When out of 1000 asperities 100 asperities
are effective in wear
= 0.01, When out of 1000 asperities 10 asperities
are effective in wear.
Also,= u . t
= π DN /60 . t
4.2.2. Abrasive Wear:
This type of wear arises from the cuttion acting of
hard surface rubbing on softer materials, as for example,
when hard surface asperities act rather like cutting tools
and remove material from softer materials. Another
example arises when loose debris of any kind is trapped
between sliding surfaces. Such debris may be extraneous,
such as sand particles, or may be the actual wear particles
created by the primary wear process.
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One method of reducing the first type of abrasive wear
is to ensure high quality of surface finish of the mating
surfaces particularly the hard surface. With modern production methods this type of wear is no longer as
serious problem. The second type of abrasive wear is more
difficult to eliminate. Suitable sealing and filtration can
reduce correct design of the contact geometry. It is often
desirable to provide grooves or other such recesses on thesurfaces of bearings, which allow the debris to ‘escape’
from the contact geometry.
Fig. 4.3 Wear due to a single conical asperity
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Fig. 4.4 Abrasive wear due to Which Olly
Consider two dissimilar surface materials in contactand is in relative motion. Assume that the hard surface
asperities remove the material from relatively soft material
surface.
Let the hard surface asperities are of conical in shape
having radius ‘a’ at the contact and height ‘h’ at the
penetration in soft material. The asperity angle is ‘θ’,
consider anyone such asperity for analysis the volume of
material to be removed from soft material in distance x is
V = The projected area of conical asperity in the
direction of motion x distance x.
= ½ x 2a x h x X but h = a. tan θ
V = a . a tan θ . x
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= a2 tan θ . x
For plastic deformation of material
W ∝ A ( area of contact)
= A . H
i.e. W = A . H = A . H/3
The area of contact in the direction of a normal load is A =
π a2 / 2
W = πa2 / 2 . H
a2 = 2W / π H
Substituting value of a2 in equation of volume
V = a2 tan θ . X
Vabr = 2W tan θ . X
πH
Here K abr = X tan θ = 6 tan θ
π π
Vabr = K abr . W.XH
Where Vadh = 1/3 W.X . K adh
H
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Vw = Vadh + Vabr = K adh – W.X + K abr + W.X
3 H H
V = K w . W.X
H
Where K w = K abr + K adh /3
In all practical cases the wear occurs due to the
abrasive and adhesive wear. Therefore the general equation
of wear can be given as
Vw = K w . W.XH
4.2.3. Fatigue Wear:
It is well known that if materials are loaded and
unloaded cyclically they exhibit fatigue failure. This type
of failure can occur after large number of loading cycles,
even though the load is less than that which we could
normally expect to produce failure in a single load
application. It is usual to express such behavior bay
logarithmic graph of stress S against the number N of
cycles to failure (the S/N curve, as it is usually called) such
as fig. 4.5. Here we see that the lower the applied cyclical
stress have the longer the life of the material.
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Fig. 4.5 A typical S-N fatigue curve
Fig. 4.5 shows a typical S-N fatigue curve. If we
consider the interaction of asperities during the sliding of
one surface over another we can see the possibility of
fatigue mechanisms being broken off the asperities to
produce wear debris. A simple experiment to illustrate this
is to run one's finger a many times along the teeth of a
comb. The teeth (asperities) are continuously being loaded
and unloaded due to repeated traversals by one’s finger and
after many cycles they ultimately break due to fatigue.
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4.2.4. Corrosive Wear:
Any clean metal surface reacts with its environment to
form contaminant films, and the rate of formation of suchmm is initially very rapid but decreases as the ‘corrosive’
film thickens. In many instances, such as oxide (rust) films
on steel, these surface films adhere only loosely to the
surface. Rubbing therefore removes the films leaving
exposed ‘clean’ metal, which immediately reacts with itsenvironment to provide new surface films, which are again
removed during ragging. So materials continuously bang
removed from the surface, and wear is taking place. The
chemistry of such reaction is beyond the scope of this book,
but fortunately is not needed for the understanding of the
basic mechanism.
A further effect of corrosive environments is to
enhance the abrasive action of wear debris. Most metal
oxides are harder than the metal itself sot that if metal
debris is created, this becomes oxidized and gives a rate of
abrasion greater than that which would otherwise occur. A
good example of this occurs with relatively soft aluminum,
where the oxidized wear debris is very hard abrasive.
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Indeed aluminum oxide is often used as the cutting agent in
girding wheels and the like.
Corrosive effects are not entirely deleterious. In thechapter on friction it has already been shown that the
presence of oxide films in preventing metal-to-metal
contact greatly reduces the coefficient of friction. In other
applications surface films are deliberately produced to
avoid metallic contact. The so-called E.P. (extreme pressure) additives to lubricating oil produce surface films
such as chlorides and sulphides and provide protective
surface layers. In a sense these could be more properly be
called extreme temperature films rather than extreme
pressure films. Their main characteristic is that chemical
stability at the high temperatures of the high pressure
contacts in such situations as hypoid gears as used in motor
car back axles.
4.2.5. Erossive Wear:
Erossive wear is mainly because of the erossion which
is a combined effect of mechanical stresses under the
ambient condition for example the wear on the bearing
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surface of a concrete mixture operated in an open
atmosphere.
4.2.6. Fretting:
This is not really a separate mechanism of wear but it
is treated separately because it arises in rather special
circumstances. It shows one particular wear process may be
a complicated combination of several mechanisms of wear,and also demonstrates the deleterious effects of any wear
debris which may become trapped in the contact system.
Fretting effects are associated with the contact of surfaces
in which the sliding motion is an oscillation of relatively
small amplitude, often only a few micrometers. Since
vibrations occur in virtually all machines we find fretting
occurring between surfaces in contact such as bolted
components, splines and components located by friction
rise to small amplitude oscillatory displacements between
the surfaces in contact.
4.3. Tribological Properfies of Plastics:
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In recent years there has been a significant growth in
the use of plastics to replace metals in many bearing
applications. In general the friction and wear of plastics can be explained by the adhesion theories already discussed.
The friction coefficients of plastics are not particularly low,
but their main advantage is that they wear at reasonably
low and predictable rates. One notable exception to this
behavior is PTFE (polytetrafluoroethylene) whose frictioncoefficient may be very low, about 0.05. This very low
friction value seems to be associated with the very low
adhesion of this material, because of this it is used in non-
stick kitchenware. It is also reasonably hard due to the
mechanical interlocking of its molecules. So it is
extensively used in bearings, where the loads and speeds
are moddest, sine its is almost self-lubricating and is highly
reliable.
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Fig. 4.6. The change in dimension due to wear.
A useful design parameter for all trips of plastics used
in bearings is the ‘pv’ factor (which is the product of thenominal contact pressure and the sliding speed.).
Consider a block of this material sliding on a metal
surface, fig. 4.6. The rate of energy dissipation against
friction is µWv. It is reasonable to assume that the volume
wear rate of the block. V = – dV / dt, is proportional to this
rate of energy dissipation, so
V ∝ µWv.
The block will therefore wear to a depth d such that
the volume wear V is given by
V= Ad.
Hence its rate
V= Ad.
Where d is the rate of increase of d with time. Thus
d = V ∝ µWv ∝ πν
A AThis shows that the rate of change of dimension of the
block is proportional to the product Pv. This is the rate of
change of bearing clearance with time in any practical
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application and as such as in more useful parameter than
the actual volume of material removed.
For any material the allowable value of the pv productmay be defined and fig. 2.7 shows a typical result of PTFE
if the wear rate is to be a dimensional change of 25 mm in
100 hours.
Plastic bearings and particularly PTFE bearings must
be operated within their approved pv ratings. These ratingsare associated with wear and are particularly subject of
thermal effects due to the decomposition of the surface.
Thus at their ambient temperatures the pv factor for such
materials is considerably reduced. Such materials when
used in practical bearing designs are often associated with a
metallic matrix which provides additional strength and
improves the thermal conductivity thus allowing the easier
escape of the heat generated in rubbing. When plastics
slide, problem can arise from the generation of electrostatic
charges. In such 10 situations, designers must incorporate
earth paths to minimize the build-up of charges.
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Fig. 4.7 A typical 'pv' curve
4.4. The Measurement of Wear:
In most engineering machinery the rate of wear is
relatively small, typically changes in dimensions of
micrometers per year. Also wear takes place in ‘real time’
and laboratory tests aye to devise conditions where the
wear processes are considerably accelerated so that results
are produced in days rather than years. In so far as such
tests are artificially accelerated their results should viewed
with some caution.
4.4.1. High-Pressure Contact Tests:
These tests produce accelerated results by applying
loads over very small areas of contact. Various geometrical
arrangements are used as shown in fig. 4.8. In each of these
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tests specimen A is the material being worn away, the
degree of wear being measured by either a change in
dimensions or a loss of mass from the material., Many suchmachines also measure the frictional force at the contact.
Fig. 4.8 Pin on Ring
Fig. 4.9 Pin on Disc
Such contact geometries can be studied either in the
ordinary atmosphere or in a totally enclosed chamber where
the atmosphere may be controlled, as to such properties as
the gaseous environment, pressure, temperature and
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humidity. A very common form of such apparatus is
designed to carry out these tests at various reduced
pressures. In a high vacuum the formation of oxide films isinhibited, so we obtain useful information on the friction
and wear of the materials themselves. UHV apparatus is
also useful for measuring the adhesion between surfaces in
contact, and we have already seen that such information is
very valuable to our understanding of friction.
Fig. 4.10 A simple crossed cylinder wear machine
More complex geometrical arrangements are shown in
fig. 4.11. The Fig. 4.11 (a) shows the four-ball arrangement
in which a rotating ball rubs against three stationary balls to
produce wear scars whose size is an indication of the
volume wear. This arrangement is extremely popular for
industrial testing since the test specimens, the balls, are
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readily available at low cost from ball bearing
manufacturers. Fig. 4.11(b) shows the so-called ‘disc
machine’ which is very useful for the study of wear under combinations of rolling and sliding. When the peripheral
speed of both discs is the same one has pure rolling whilst a
difference in speeds implies some additional sliding. This
rather complex roll/ slide process often occurs in
machinery, perhaps the best example being the contact between meshing gear teeth depicted in fig. 4.12. At the
initial contact we have rolling and sliding between the teeth
which becomes pure rolling at the pitch point B followed
by rolling with sliding in the opposite sense during the arc
of disengagement.
Fig. 4.11 The four ball wear machine and disc on disc
wear machine
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A. Initial contact - rolling and sliding
B. Pitch point contact-pure rolling
C. Final contact-rolling and sliding
Fig. 4.12 The progress of contact between gear teeth
( Solved Problems )
EX.1. A pin on disc experiment the rate of vol. of wear
observed was 0.1 mm3/min. Pin was placed at 30 mm
radial distance from the centre of disc. The disc was
rotated at 500 rpm. Load on pin is 20 N. Material of pin
is brass, hardness is 120 N/mm2. Material of disc is steel,
hardness 280 N/mm2. Determine the wear constant.
Solution :
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Data given:
V = 0.1 mm3/min
r = 30mm N = 500 rpm
W = 20N
H p (pin) = 120 N/mm2
Hd (disc) = 280 N/mm2
V = K w . W.2π rNt
H p
V / t = Wear rate = 0.1 = K w . 20 x 2 x π x 30 x 500
120
K w = 6.3662 x 10-6
Ex.2. Determine the wear constant K w by using a pin on
disc wear measuring test ring. The testing data are as
follows:
(1) Pin diameter at the tip 3.0 mm.
(2) The maximum diameter of the conical pin =
10.0 mm
(3) The cone height of the pin is = 9.0 mm.
(4) Load on the pin = 50 N
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(5) The disc diameter = 150 mm.
(6) The pitch radius at pin contact = 60 mm.
(7) r.p.m. of the disc. = 600.(8) The hardness of the pin = 20 N/mm2.
If the experiment is conducted for 130 hours. The
wear reading show the linear in unit of the wear
volume. The final diameter of the pin was 8.0 mm. The
material of the pin is brass and the material of the discis hardened steel. .
If the above tested material is required to use in a
machine for shaft and bearing. Determine the life of the
bearing for a shaft rotating at 140 r.p.m. under a load of
10N. The bearing diameter is 40mm. The maximum
allowable radial wear is 1.0 mm. The length to diameter
ratio of the bearing is 1.0.
Solution:
tan θ = (d2 – d1)/2
r = (10 – 3)/2 = 0.3888
9
θ = 21.25°
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K w = 5 (dt3 – d1
3)H
4 tan θ W.R . N.T
t in sec:
Time require = 130 hours
= 130 x 60 x 60 secs.
t = 468000 secs.
K w = 5 (83 – 33)20
4 tan 2125 50 x 60 x 600 x 468000
= 3.7013 x 10-8
Ans = I
Now N = 140 rpm
K w = 3.7013 x 10-8
W = 10 N
D = 40 mm
L = 40 mm
r w = 1.0 mm
Life of the bearings t = (D) . r w . L . H
K w.W.πDN/60
= (40) x 1.0 x 40.20 x 603.7013 x 10-8 x 10 x π x 40 x 140
= 1.47428 x 108 sec. x 2
= 40952.135 hours x 2
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= 81904.3 hours.
Ex.3. In an experiment of wear with the plastic bearing
it is found that for the change in dimension of 0.1 mmunder the load of 1.5 N in a slender bearing. The
velocity required is m/sec. For the same operating
conditions, if the load and speed are doubled than what
will be the change of dimensions?
Data given:dt = 0.1 mm
W = 1.5 N
u = 1 m/sec.
dd1 = µ W1 u ,
dt1 A1
&
dd2 = µ W2 u2
dt2 A2
But
A1 = A2
W2 = 2 W1
u2 = 2 W
dt1 = dt2
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dd2 = P2 u2
dd1 P1 u1
dd2 = 2.2
0.1
dd2 = 0.4 mm
For the same operating conditions change of
dimension is 0.4 mm if load & pressure both (velocity) are
doubled.
Ex.4. In a crossed cylinder wear measuring experiment
the stationary specimen is brass in contact with a
rotating cylinder made all of hardened cylinder having
radius 50 mm and thickness 25 mm. is rotating at 500
rpm. The rate of Wear was observed to be constant is
found 2mm in 60 hours of a constant operation under a
50 N load.
Determine the wear const. if the hardness of brass
is 120 N/mm2
Solution:
Data given:
H = 120 N/mm2
W = 50 N
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R = 50 mm
L = 25 mm
N = 500 rpmt = 60 hours
h = r w = 2 mm
cosθ = R-h = 50.2 = 0.96
R 50
θ = 16.26°
K w = R 2 (θ – sin 2θ / 2) L H
W (π D N / 60) t
t = 60 x 3600 secs.
K w = 502 [16.26 x π/180 – sin {(2 x 16.26 )/ 2}] 25 x
120
50 x [(π x100 x 500)/ 60] x 60 x 3600
= 3.977 x 10-6 Ans.
Ex.5. A 30 mm long brass bearing is used to support a
steel shaft having 50 mm diameter, and a steady radial
load of 60 N. Shaft is rotating at 500 rpm. Shaft surface
is having an average asperity angle 10°. The hardness of
the bearing material is 200 N/mm2. Determine the time
required to exceed the radial wear 2 mm.
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Solution:
Data given:
r w = 2 mmL = 30 mm
D = 50 mm
W = 60 N
N = 500 rpm
θ = 10°.
H = 200 N/mm2
(Questions)
1. Write short notes on the following:
(i) Fatigue wear
(ii) Pin on disc wear measurement.
2. Explain cross cylinder wear machine.
Define ‘Wear’. State the condition where ‘wear’ proved
to be boneficial.
3. In pure abrasion wear, show that the wear quantity6 W.X
V = 6 tanθ WX
π H
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Where θ = average asperity angle.
W = applied load.
X = sliding distance.H = Hardness of material
4. Explain four ball Wear machine.
5. Classify the wear mechanisms. Explain the methods to
eliminate the wear, of each mechanism.
6. Define fatigue Wear.
7. State the mechanisms of wear. Describe the methods to
eliminate wear.
8. Show that the change in dimension due to wear for the
case of plastic - metal contact is proportional to the
product of pressure and the relative velocity of the
contact surfaces.
9. Define abrasive wear.
10. Define wear. Why do you consider that wear is
beneficial in running in and planned obsolescence?
11. Explain elimination of wear.
12. Show that the volume of wear due to adhesion and
abrasion is
Vw = K w W.x
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H
Where K w = wear constant.
13. A steel shaft runs within PTFE bearing at 10 m/sec.After 1000 operating hours the bearings were warn 2 mm
radially. Find out the load on each bearing if the contact
area of each bearing is 0.01 m2. The wear constant may
be taken as 10-21 m2/N.
Suggest the various possibilities to improve the performance of such bearing.
14. In a cross cylinder wear measuring experiment the
stationary specimen is brass in contact with a rotating
cylinder made out of hard steel; having radius 50 mm
and thickness 40 mm is rotating at 960 rpm. The size of
the stationary specimen is 100 mm length, 25 mm width
and 25 mm height. The rate of wear was observed to be
constant, 100 N load. Find out the wear constant K w.
If above tested material is required to use in a machine
for shaft and bearing. Determine the life of the bearing
for a shaft rotating at 250 rpm under load of 25 N. The
maximum allowable radial wear is 0.5 mm and length of
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the bearing is 75 mm. Take hardness of the brass = 120
N/mm2.
15. Show that the life of a sleeve bearing due to abrasivewear is
L = b.l.H hrs.
1800 kwWw
Where b is the permissible radial wear of sleeve bearing
l is the length of the sleeve bearing, m
H is the hardness of sleeve bearing, N/m2.
hw is the wear coefficient.
W is the radial load, N
ω is the angular velocity rad/sec.
16. In an experiment on a pin-on-disc test rig the
following observation were made:
(i) Diameter of the pin and its material: 10 mm, Brass
(ii) The pitch diameter of the disc and its material: 120
mm Brass.
(iii) The change in length of the cylindrical pin due towear in 150 hours: 5 mm
(iv) The rotational speed of the disc is: 500 rpm.
(v) The hardness of the pin is: 20 N/mm2
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(vi) Load on pin is: 200 N.
If the above tested material is to be used in a machine for
shaft and bearing, determine the life of the brass bearingfor a steel shaft rotating at 150 rpm. under load of 50N.
The permissible radial wear is 1.0 mm. The length/
diameter ratio of the bearing is 1.0. The length of the
bearing 50 mm.
17. In a pin on disc experiment, the test piece pin was madeout of brass and disc was made out of stainless steel. The
pin was located at 50 mm radial distance from the disc
centre. The asperity angle on the disc surface was 3°.
The power required to rotate the disc at constant speed
960 rpm was 500 watts. The testing data are as follows:
Pin diameter at tip = 3.0 mm
The maximum diameter of conical pin = 12.0 mm
The cone height of pin = 12.0 mm.
The hardness of pin material = 20 N/mm2
The wear constant = 3.0 x 10-8.
The wear reading shows the linear increment of the wear
volume. Determine the time required for the experiment
of wear if the final tip diameter of the pin is 10.0 mm.
8/2/2019 Tribo Notes
http://slidepdf.com/reader/full/tribo-notes 37/37
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